Câu 19 trang 29 Sách bài tập (SBT) Toán 8 tập 1

Dùng quy tắc đổi dấu để tìm mẫu thức chung rồi thực hiện phép cộng:

a. \({4 \over {x + 2}} + {2 \over {x - 2}} + {{5x - 6} \over {4 - {x^2}}}\)

b. \({{1 - 3x} \over {2x}} + {{3x - 2} \over {2x - 1}} + {{3x - 2} \over {2x - 4{x^2}}}\)

c. \({1 \over {{x^2} + 6x + 9}} + {1 \over {6x - {x^2} - 9}} + {x \over {{x^2} - 9}}\)

d. \({{{x^2} + 2} \over {{x^3} - 1}} + {2 \over {{x^2} + x + 1}} + {1 \over {1 - x}}\)

e. \({x \over {x - 2y}} + {x \over {x + 2y}} + {{4xy} \over {4{y^2} - {x^2}}}\)

Giải:

a. \({4 \over {x + 2}} + {2 \over {x - 2}} + {{5x - 6} \over {4 - {x^2}}}\) \( = {4 \over {x + 2}} + {2 \over {x - 2}} + {{6 - 5x} \over {\left( {x + 2} \right)\left( {x - 2} \right)}}\)

\(\eqalign{  &  = {{4\left( {x - 2} \right)} \over {\left( {x + 2} \right)\left( {x - 2} \right)}} + {{2\left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x - 2} \right)}} + {{6 - 5x} \over {\left( {x + 2} \right)\left( {x - 2} \right)}} = {{4x - 8 + 2x + 4 + 6 - 5x} \over {\left( {x + 2} \right)\left( {x - 2} \right)}}  \cr  &  = {{x + 2} \over {\left( {x + 2} \right)\left( {x - 2} \right)}} = {1 \over {x - 2}} \cr} \)

b. \({{1 - 3x} \over {2x}} + {{3x - 2} \over {2x - 1}} + {{3x - 2} \over {2x - 4{x^2}}}\) \( = {{1 - 3x} \over {2x}} + {{3x - 2} \over {2x - 1}} + {{2 - 3x} \over {2x\left( {2x - 1} \right)}}\)

\(\eqalign{  &  = {{\left( {1 - 3x} \right)\left( {2x - 1} \right)} \over {2x\left( {2x - 1} \right)}} + {{\left( {3x - 2} \right).2x} \over {2x\left( {2x - 1} \right)}} + {{2 - 3x} \over {2x\left( {2x - 1} \right)}}  \cr  &  = {{2x - 1 - 6{x^2} + 3x + 6{x^2} - 4x + 2 - 3x} \over {2x\left( {2x - 1} \right)}} = {{1 - 2x} \over {2x\left( {2x - 1} \right)}} = {{ - \left( {2x - 1} \right)} \over {2x\left( {2x - 1} \right)}} = {{ - 1} \over {2x}} \cr} \)

c. \({1 \over {{x^2} + 6x + 9}} + {1 \over {6x - {x^2} - 9}} + {x \over {{x^2} - 9}}\)\( = {1 \over {{{\left( {x + 3} \right)}^2}}} + {{ - 1} \over {{{\left( {x - 3} \right)}^2}}} + {x \over {\left( {x + 3} \right)\left( {x - 3} \right)}}\)

\(\eqalign{  &  = {{{{\left( {x - 3} \right)}^2}} \over {{{\left( {x + 3} \right)}^2}{{\left( {x - 3} \right)}^2}}} + {{ - {{\left( {x + 3} \right)}^2}} \over {{{\left( {x + 3} \right)}^2}{{\left( {x - 3} \right)}^2}}} + {{x\left( {x + 3} \right)\left( {x - 3} \right)} \over {{{\left( {x + 3} \right)}^2}{{\left( {x - 3} \right)}^2}}}  \cr  &  = {{{x^2} - 6x + 9 - {x^2} - 6x - 9 + {x^3} - 9x} \over {{{\left( {x + 3} \right)}^2}{{\left( {x - 3} \right)}^2}}} = {{{x^3} - 21x} \over {{{\left( {x + 3} \right)}^2}{{\left( {x - 3} \right)}^2}}} \cr} \)

d. \({{{x^2} + 2} \over {{x^3} - 1}} + {2 \over {{x^2} + x + 1}} + {1 \over {1 - x}}\)\( = {{{x^2} + 2} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {2 \over {{x^2} + x + 1}} + {{ - 1} \over {x - 1}}\)

\(\eqalign{  &  = {{{x^2} + 2} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{2\left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{ - \left( {{x^2} + x + 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}  \cr  &  = {{{x^2} + 2 + 2x - 2 - {x^2} - x - 1} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {{x - 1} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {1 \over {{x^2} + x + 1}} \cr} \)

e. \({x \over {x - 2y}} + {x \over {x + 2y}} + {{4xy} \over {4{y^2} - {x^2}}}\)\( = {x \over {x - 2y}} + {x \over {x + 2y}} + {{ - 4xy} \over {\left( {x + 2y} \right)\left( {x - 2y} \right)}}\)

\(\eqalign{  &  = {{x\left( {x + 2y} \right)} \over {\left( {x - 2y} \right)\left( {x + 2y} \right)}} + {{x\left( {x - 2y} \right)} \over {\left( {x - 2y} \right)\left( {x + 2y} \right)}} + {{ - 4xy} \over {\left( {x - 2y} \right)\left( {x + 2y} \right)}}  \cr  &  = {{{x^2} + 2xy + {x^2} - 2xy - 4xy} \over {\left( {x - 2y} \right)\left( {x + 2y} \right)}} = {{2{x^2} - 4xy} \over {\left( {x - 2y} \right)\left( {x + 2y} \right)}} = {{2x\left( {x - 2y} \right)} \over {\left( {x - 2y} \right)\left( {x + 2y} \right)}}  \cr  &  = {{2x} \over {x + 2y}} \cr} \)

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